Sheet 6
Create the directory sheet06 and make it your working directory.
Example 1. This example is warm-up to learn to use Matlab’s DFT algorithm and generate
plots.
a) Plot the absolute value of the DFT yb = {b
yk } of f (x) = x + x2 , x ∈ [0, 2π] as follows:
n = 2∧5;
x = linspace(0,2*pi,n);
w = 1:length(x);
y = x + x.∧2;
yt = fft(y);
plot(w,abs(yt),’o’)
xlabel(’k’);ylabel(’Modulus of DFT’)
Try using n = 26 and 27 .
Repeat this for :
b) f (x) = x sin(10x)e−x , x ∈ [0, 2π]
c) f (x) = cos(10x) sin(10x)e−x , x ∈ [0, 2π]
d) f (x) = sin(30x) cos(10x), x ∈ [0, 2π]

0≤x≤π
 −1 if
e) f (x) =

1 if π < x ≤ 2π
Example 2. In this example, you are going to plot the modulus of the Discrete Fourier
Transform of f (x) = (x−π)2 , x ∈ [0, 2π] along with the n times the modulus of the complex
Fourier coefficients to see how they relate to one another.
Proceed as follows. First enter the function and give the desired n
clear
f = @ (x) (x-pi).∧2;
n = 2∧3;
Find the complex Fourier coefficients βk :
syms x k;
b k = int(f(x)*exp(-i*k.*x),x,0,2*pi)/(2*pi)
Remember to execute these lines, then copy and paste the output from the command window
in the following line:
kth coef = @ (k) paste here
Now compute the coefficients:
for m = 1:n-1
b(m) = int(f(x),x,0,2*pi)/(2*pi);
end
b 0 = kth coef(0);
List all the coefficients (the double converts to decimal form)
coeffs = double([b 0,b(1:n-1)]);
Find the DFT
x = linspace(0,2*pi,n);
y = f(x);
yt = fft(y);
Finally, plot the coefficients and the DFT on the same graph:
w = 1:length(x);
plot(w,abs(yt),’o’,w,n*abs(coeffs),’r’,’linewidth’,2)
xlabel(’k’);
ylabel(’Modulus of DFT and n times Complex Fourier Coefficient’)
legend(’n|DFT|’,’|Fourier Coefficient|’)
Sometimes it is useful to zoom in on what is going on near the x−axis. Remember you do
this with an axis statement, for example in this case, axis([0 n 0 10]) would be suitable.
Try different values of n, say 24 and 25 .
Aside: Recall if βk is the complex Fourier coefficient and yb = {b
yk } is the DFT, then we
have the approximation ybk ≈ nβk for k small compared to n. You should see how this is
working as you change the value of n.
Example 3. Repeat for the following (I suggest you save each separately: ex3a.m, ex3b.m,
etc.). Again, vary n to see how the approximation changes and make appropriate axis
statements to zoom in on the action near the x−axis, if necessary.
a) f (x) = x + x2 , x ∈ [0, 2π]
b) f (x) = xe−x sin(10x), x ∈ [0, 2π]
Note that when we are plotting |βk | as a function of k = 1, . . . , n, any peak in the plot at
some value k = k0 corresponds to a large frequency component for the frequency ω = k0 .
Because of the sine that appears in f (x), wiggles in the plot of f will occur with an approximate frequency of ω = 10. So you should see a large frequency component in the DFT plot
at k = 10. Do you?
c) f (x) = cos(10x) sin(10.2x), x ∈ [0, 2π]
In this example there should be two peaks corresponding to ω = 0, 20 in the first half of the
plot of the DFT.
d) f (x) = e−3x sin(30x) cos(10x), x ∈ [0, 2π]
There should be two peaks in the first half of the DFT. What are the corresponding values
of the frequency ω ?
e)
f (x) =

 −1

1
if
0≤x≤π
if
π < x ≤ 2π
Hint: On this one, you will need to modify the method a bit. In the lines defining b k
and b 0, enter the integral in two pieces corresponding to [0, π] and [π, 2π] and enter the
corresponding value of f in place of f (x).
Sheet 6: Further Exercises
f)
f (x) =



π 2 − x2
if
0≤x≤π
−π 2 + (x − π)2
if
π < x ≤ 2π